text:
Delays between "50", "40", "30", "20", and "10" cockpit announcements
When a large airplane is about to land, the following announcements are heard in the cockpit:
"Fifty!"
(delay)
"Forty!"
(delay)
"Thirty!"
(delay)
"Twenty!"
(delay)
"Ten!"
What are the delays' values that would indicate that the plane is descending at a normal rate?

You understand that these are calling out AGL altitudes right? The delay depends on the descent rate. "Normal" is a matter of aircraft and the situation.

Quickly plot them out and verify that they are asymptotically approaching zero.  :-)

That implies an infinitely long runway.

@RonBeyer, the aircraft is Boeing-777, and "normal" situation is a standard 3 or so degree glideslope. I thought the delays would depend on the angle but not on aircraft.

It depends on aircraft because even though they use the same 3-ish degree glideslope, the aircraft have different speeds. For an extreme example, I can follow a 3 degree glideslope in a Piper Cub at 50 knots, or a 777 at 150 knots, it's the same path, but those call-outs (if the cub had them) would happen a lot faster on a 777.

Are you expecting the pilot to have a stopwatch to time the delays to ensure the rate of descent is appropriate instead of relying on the array of instruments already in front of him?

That depends on a lot of things. That’s just a radar altimeter counting down on short final.  By that point almost all of the control of the rate of descent is being done by the pilot vis a vis visual cues to arrive at a particular touchdown area on the runway.  Pilot will increase or decrease their rate of descent accordingly.  Delays or lack there of in that aural warning is irrelevant.  Pilots do however use such audible cues as signals for reducing engine thrust and pitching to the appropriate roundout attitude for touchdown, e.g. the ‘RETARD RETARD’ cue.

There is no value, or set of values that is "correct", but for perspective if the approach were flown at a reasonable 600 feet per minute rate of descent the plane would be descending at 100' every 10 seconds, or 10' every second.  (No aircraft type, airspeed, or glideslope angle calculations are necessary - although they affect rate of descent, the descent rate is the only factor in play for the interval of these calls.)
So, if the pilot did nothing to arrest the rate of descent the 10' increment calls you gave in your example would come every second in a continuous conversational cadence all the way to touchdown.  The goal of a butter soft landing would be for the rate of descent to equal zero just as the wheels touch down, but a "normal" time interval between calls is difficult to quantify because it isn't linear.
I haven't personally flown with an automated altitude alerting system such as the one you describe, nor have I had a copilot making such calls, (heck, a solid majority of my landings have been at 600fpm...) but just like the visual cues a pilot will use to judge height above touchdown to flare, I would imagine that you would quickly get a feel for the sense of timing and appropriate decreasing pace of the interval between calls, and learn to adjust intuitively with some practice.
i.e. Inside of 50 feet if the calls start coming every second the hair on the back of your neck ought to tingle, and you will need to quickly think about starting to apply some back pressure to arrest the descent and prevent a hard landing.


text:
Will the autopilot raise the AoA above the critical AoA to maintain altitude?
NOTE: This question is regarding commercial airliners (eg: Boeing 737, Airbus A320, etc.. This is for any given configuration of the aircraft)
Will the autopilot raise the AoA above the critical AoA to maintain the given altitude or the glideslope(if intercepted)? For example, both engines failed in the case of BA-38 the autopilot tried to stay on glideslope which resulted in a speed loss. (thanks to the pilot's timely action to disconnect the autopilot which avoided it from stalling out and losing too much speed which saved the lives of many people).
Also, will the autopilot reduce the AoA on its own if the aircraft stalls out?

Ralph wrote a nice answer for the Boeing 737, which is a conventionally controlled aircraft, meaning the yokes are mechanically linked to the flight control actuators.
Airbus A320
The Airbus A320 however is a fly-by-wire controlled aircraft. This means that both sidestick input and autopilot commands are sent to flight control computers, which then calculate how to move the control surfaces to achieve the desired flight path:

(Airbus A320 FCOM - Flight Controls - Description)
While the autopilot will happily command inputs that would increase the angle of attack above its critical value, the flight control computers have built-in protections against this. These protections are active regardless of whether the pilots control the aircraft via the sidesticks or the autopilot controls the aircraft:

Control with autopilot engaged:

The ELACs and SECs limit what the autopilot can order. [...]
All protections of normal laws remain effective.

(Airbus A320 FCOM - Flight Controls - Normal Law)
In the case of increasing angle of attack, the following protection law would apply:

High angle of attack protection
Under normal law, when the angle of attack becomes greater than αprot,
the system switches elevator control from normal mode to a protection
mode, in which the angle of attack is proportional to sidestick
deflection. That is, in the αprot range, from αprot to αmax, the
sidestick commands α directly. However, the angle of attack will not
exceed αmax, even if the pilot gently pulls the sidestick all the way
back. If the pilot releases the sidestick, the angle of attack returns
to αprot and stays there.
This protection against stall and windshear has priority over all
other protections. The autopilot disconnects at αprot + 1°.

(Airbus A320 FCOM - Flight Controls - Normal Law)
Since the autopilot disconnects itself at αprot + 1°, it would not automatically recover from a stalled condition. However, AoA would reduce now anyway because the flight control computers would not allow any further elevator up actuation until AoA is below αmax again.
In addition to limiting the angle of attack via the elevator, the aircraft will also automatically set maximum thrust at some point. This is called Alpha-Floor protection:

Alpha-Floor protection
Alpha-floor protection automatically sets the thrust at TOGA thrust,
when the aircraft reaches a very high angle of attack.
The Flight Augmentation Computer (FAC) generates the signal that
triggers the alpha-floor mode. This, in turn, sets TOGA thrust on the
engines, regardless of thrust lever position (Refer to 1.22.30 A/THR
modes).
The FAC sends this signal when the angle of attack is above a
predetermined threshold, that is a function of the configuration.

(Airbus A320 FCOM - Auto Flight - Flight Augmentation)
Boeing 777
The accident you linked was in a Boeing 777, which is also fly-by-wire controlled and therefore also provides flight envelope protections. However, Boeing uses a somewhat different approach, where the protections increase the required force on the yoke to overcome them, rather than limiting what is being transmitted to the flight control actuators. This means, pilots can always overpower the envelope protections in a fly-by-wire Boeing.
The autopilot will however obey the flight envelope protections and therefore not stall the aircraft:

Flight Envelope Protection
There are three forms of flight envelope protection in the autopilot:

stall protection
overspeed protection
roll envelope bank angle protection

An AUTOPILOT caution message and roll or pitch mode failures alert the
pilot if the envelope is exceeded, and the autopilot prevents further
envelope violations.

(Boeing 777 FCOMv2 4.20.19 - Automatic Flight - System Description)
The stall protection will not allow the autopilot to trim the aircraft into a stalled condition and — similar to the Airbus Alpha-Floor protection — also use the autothrottle to increase airspeed:

Stall Protection
Stall protection reduces the likelihood of inadvertently exceeding the
stall angle of attack by providing enhanced crew awareness of the
approach to a stall or to a stalled condition.
Stall protection limits the speed to which the airplane can be
trimmed. At approximately the minimum maneuvering speed, stall
protection limits the trim reference speed so that trim is inhibited
in the nose up direction. [...]
The autothrottle can support stall protection if armed and not
activated. If speed decreases to near stick shaker activation, the
autothrottle automatically activates in the appropriate mode (SPD or
THR REF) and advances the thrust to maintain minimum maneuvering speed
(approximately the top of the amber band) or the speed set in the mode
control panel speed window, whichever is greater. The EICAS message
AIRSPEED LOW is displayed.

(Boeing 777 FCOMv2 9.20.12 - Flight Controls - System Description)
In the accident flight, autothrottle inputs where obviously no longer available since both engines had already failed. This also has other implications (e.g. both engine generators fail, which requires emergency electrical power from battery or RAT), which can result in the loss of flight envelope protections:

Flight Control System Secondary Mode
When the PFCs can no longer support the normal mode due to internal
faults or lack of required information from other airplane systems,
they automatically revert to the secondary mode. [...]
The following functions are not available in the secondary mode:

autopilot [...]
envelope protection [...]

(Boeing 777 FCOMv2 9.20.7 - Flight Controls - System Description)

On the 737, some modes will hold altitude (or glideslope) no matter what, while other modes will change over (or "revert") to a descent to maintain a minimum speed.
It's probably a long discussion regarding why "this" mode will revert to a constant IAS descent while "that" mode won't; the easiest case is probably GS Capture: it will track the glideslope NO. MATTER. WHAT.  It will not give up the glideslope and allow the aircraft to sink below it unannounced, because that can kill you without warning. If it holds the GS and lets airspeed decay (i.e. with power at idle & not coming up as it should), hopefully the stall warning will get your attention & you add power to avoid the stall and the crash. Obviously, being in that situation, low to the ground & airspeed decreasing, is a place you never want to be, but the engineers apparently say that there, giving up the GS to hold airspeed is even more dangerous than doing the opposite.
Altitude Hold has the same priority, for slightly different reasons. Vertical speed and most VNAV modes, do the opposite, and will revert to a minimum speed rather than flying the aircraft into a stall. (Those modes also have speed reversions at the high end - they will revert to hold VMO or MMO at the expense of the selected V/S or path, rather than holding the latter & overspeeding the aircraft. Alt Hold & GS Capture have no speed reversions high or low & in those modes you could overspeed things with power pushed up long enough.)
Vertical Speed mode is probably easiest to explain here; if I dialed in a 2,000 feet/minute climb and forgot about it, the safest course is probably to continue the climb, at a reduced vertical speed but a safe flying airspeed, rather than stall (probably pretty violently) by pulling harder & harder to hold a climb rate that I no longer have the power (at high altitude) to sustain. So the V/S mode does have a min speed (or, perhaps, an AoA) reversion, where Alt Hold and GS Capture do not.

The Rockwell Collins A/P as used on the CRJs would always just disconnect itself when you approached shaker, whatever mode it was in.  If you were in ALT and pulled the thrust to idle, as the barberpole came up the A/P would kick off with the warbler going off, and then the plane would pitch down to maintain trim speed on its own if you didn't intervene, although if you were at high altitude you'd want to help it along.

@JohnK Many different ways to skin the cat, obviously, and probably some differing philosophies behind some of them. Just curious, what did the FD command in a case like you describe? Holding altitude regardless? Or did they retract at the same time the AP disengaged?

Actually I can't remember if the FD also went out of view or it went into pitch hold, or stayed in ALT, or what.  My last sim session was in 2012.  I'd have to look it up.

source: Stack Exchange

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sort: votes


text:
What 's the difference between Angle of Attack and Pitch?
I am interested in knowing the theoretical and practical differences between the Angle of Attack (AoA) and Pitch:

Graphical definitions
How are they related?
In what circumstances these angles are equal?
While flying, do you need to consider both or only the Pitch?

Thanks in advance :)

Angle of attack is the angle between the relative wind and the chord of the wing.  Pitch is the angle between the longitudinal axis of the aircraft and the horizon.
They are related, (and both are controlled with the elevator) but independent of each other.  In high speed straight and level flight they are probably the closest to being equal, but generally there will be some positive angle of attack needed to maintain lift.  This is due to angle of incidence; the designed-in angle between the longitudinal axis and the wing chord.
Pitch and AoA can be vastly different though.  Consider a loop:  The pitch will change 360 degrees while the AoA may stay exactly the same.
A pilot needs to consider both.  Decelerating in straight and level flight will require gradually increasing AoA to maintain lift.  (which will also increase pitch...)  At some point you will reach the critical AoA where airflow separation leads to a stall.
This can also happen in a climb or descent, and at varying pitch angles.  I will stop there and not go more into stalls or accelerated stalls because it is beyond the scope of your question, but I wanted to at least touch on it because if you understand that then the relationship between pitch and AoA will become more clear.

The reference of pitch is the horizontal. The reference of AoA is the direction of the relative wind.

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text:
Can a tiltrotor fly safely with one engine?
I am curious about tilt-rotor feasibility (such as AW609) against fixed-wing and rotor-blade aircraft (conventional helicopters).
One important issue that immediately arises in my mind is that both propellers must be highly robust against failures. This is translated in a significant disadvantage against traditional aircraft designs since:

Fixed-wing which are equipped with more than one engine have redundancy, or at least they can glide long enoughg if only equipped with one engine
Rotor-blade can still perform an emergency landing in the event of a main engine or tail-rotor failure, as this question points.

So the big question is: Can a tilt-rotor fly safely (or at least perform an emergency landing) with a faulty engine?
P.D.: For the sake of the question let's presume that the engine has not blown up or has suffered from structural integrity failure (e.g. one blade is missing).

From your linked wikipedia article: In the event of a single engine failure, either engine can provide power to both proprotors via a drive shaft; the AW609 is also capable of autorotation., so it seems that all the information given for the V-22 is applicable to the AW609.

This concern was one of the major design challenges in the development of tilt-rotor VTOL craft.  In every flying example I'm aware of, it was solved one of two ways: either the engines were embedded in the fuselage and power was transmitted through a common gearbox (much like a twin engine helicopter), or the engines in the wingtip pods were connected with a shaft that ran through the fuselage from one engine's gearbox to the other, including overrun clutches, so either engine could drive both rotors.

V-22 as an example tilt rotor.  You can see the shafts that go from the engines to the central gear box; either engine can drive both prop-rotors.
Of course, there's still the question of whether a single engine has the power to perform a hover landing.  From reading, I gather than there are "autorotate" options for these craft -- forward speed being used to store momentum in the prop-rotor system, which can then be tilted to provide lift and clear the blades from the ground for landing.  This option will be improved if there is still power from a single engine; partial tilt can be used to add lift and slow landing speed.

The facts on the autorotation characteristics for the V-22 Osprey are that an autorotation is flyable, but the rate of descent is really high (something like 4000 FPM) and so the general training is that you practice the auto in the sim, but most twin engine loss scenarios look more like a fixed wing ditching than an autorotation. (I have an old copy of a flight manual for the V-22 somewhere in a file ... )

I dropped in a picture and a brief explanation.

Thanks, @KorvinStarmast.

Short Answer: yes. If they could not, they'd likely not get certified
One engine can drive both prop rotors.

After digging into an old version of the V-22 flight manual, I find a whole pile of tables in the chapter that cover single engine flight.   Gross weight and DA and flight configuration informs what the "best" airspeed is for single engine flight.
The general approach for single engine flight to a safe landing is to fly at a particular configuration and airspeed, and then to set up the aircraft for an approach to a run-on-landing "between 55 and 65 knots."  As I understand the procedures, that final approach will have the nacelles tilted up far enough to prevent the tips of the prop-rotors from striking the ground once the aircraft is on the ground.
Flight Safety Note: I am not going to replicate (cut and paste) these procedures here for a number of reasons, one of which is that I am referring to an out-of-date manual, and some of the details may have changed.  I suspect that the current procedures are similar.
Note on the source:
An old version in the A1-V22-NFM-000 series; Part V/Chapter 12; Emergency Procedures. (There will be a similar section in the Dash Ten for USAF CV-22's).
Where a run-on-landing is not achievable, the manual calls for an approach to ...

... maintain the highest suitable air speed until deceleration is required for a no-hover touch down.

This general method is very similar to the single engine procedures that we had in the SH-2F and the SH-60B helicopters that I flew from ships in the Navy.  The aircraft's Gross Weight and the Density Altitude of the day will lead to some variations in the "best" airspeed for this; there are tables in the manual (and usually in the pocket checklist) to determine the best airspeed to fly.
Then ...

A shallow approach into the wind maintaining 30 KCAS until right before landing is recommended.

This is a little bit different from how we did this, and taught this, in those two helicopter models; we would try for a constant angle of descent approach (not a shallow approach) to a spot if we, as was usually the case, did not have single engine hover capability.  The resulting landing would be fairly firm, but that was by design.
Bottom Line
Not only is single engine flight possible, as well as single engine approach and landing in a tilt rotor, it's a well known flight characteristic that has well established procedures to accomplish.  (At least for the V-22.  The AW-609 doubtless has similar procedures in its flight manual).

Wow, that's a massive improvement over my answer.  At least if you like to know the details...  ;)

@ZeissIkon Your answer is correct, and concise. :)


text:
MIL-PRF-5606J, What does "J" means? Does it equal to MIL-PRF-5606?
MIL-PRF-5606J, What does "J" means? Does it equal to MIL-PRF-5606?
Please let me know it they are same specification or not.

J is the revision, as noted. If no letter is used, it is generally in the context of "use the latest revision", or "use the revision in place at time contract award".

I don't know this specification, but the letter at the end indicates the revision number of the document. J would be the 10th revision/ 11th edition.
Note that there are often also intermediate publications with minor improvements called 'Amendments'
Version J was published in March 2018.
The version before, H, was published in June 2003 with Amendments in 2003, 2005 and 2006, and an additional notice in 2011.
(source)
Referring to MIL-PRF-5606 can thus mean the first edition of the specification, or it can be a generic reference where the edition is omitted.
It depends on the context, but since the original publication must have been ages ago, my bet is on it being a generic reference.

Related


text:
Capturing of false glide-slope
I am curious to know a couple of things regarding glide-slope:

How often do ATCs encounter flights capturing false glide-slopes?

What is the response expected from ATCs in such a scenario? In other words, is it the duty of ATC to alert the crew of the possibility of a false glide-slope capture.

It might be quite rare.  Only two results in an NTSB search, and even those aren't really about that.

Thank you. I have come across this for the first time in the context of PIA 8303. The link is: https://www.airlineratings.com/news/completely-unacceptable-approach-preceded-pk8303-crash/

On an ils plate there is a table that computes expected altitude at certain distances from a dme ( usually in 1 mile intervals ). Pilots are expected to compare actual altitude with computed one at least 2 times .

Agreed. But as in the case of above (PIA 8303) it is a possibility, as was noted in an earlier paper. As this involves landing, not always the crew may have time for re-computation. See: https://www.researchgate.net/publication/314141130_A_global_safety_deficiency_False_glide_slope_capture_affecting_aircraft

I realised that my previous comment is wrong , those are specific actions taken by an airline .
The incident referred in your link is this : http://avherald.com/h?article=4641efec/0000&opt=0 .
" The revised policy includes a new horizontal landing gate for ILS interception from above; being the earliest of the following: the FAF, 5 DME, 4 NM for VMC operations and 5 NM for IMC operations."

in 10 years of working at a small airport serving commercial airliners never.
ILS is a pilot interpreted landing aid, ATC has no obligation to follow the plane once established on ILS. ATC has the responsibility to vector the plane to such a position and altitude such that the plane can intercept the localiser and glide slope. At airports doing dependant ILS approaches ( 2 aircraft on 2 parallel runways ) ATC follows the distance between planes, not the altitude/glide slope. If providing radar services the ATC vectors the plane until established on the localiser then it's the crews responsibility to follow the ILS procedure.

At least in the USA, however, ATC is responsible for vectoring aircraft to intercept the glideslope outside of the GS-capture point at an altitude not above the glideslope itself (nor below the published minimum intercept altitude). JO 7110.65 5–9–1c.

Thank you. Could you please share which country you are referring to?

I will reword my answer in order to be more clear.

Thank you @cavver


text:
Why does V2 decrease for intersection takeoff
I have noticed that the VR and V2 decreases for same FLEX when departing from an intersection runway even when sufficient stop margin is available. What is the reason for this? And also does EFB calculate takeoff performance based on balanced field?

I suspect the issue is the climb gradient required for obstacle clearance assumptions.  the intersection takeoff moves my point of liftoff down the down the runway by some number of feet.  A full runway V2 may be based on me being off 3,000 feet before the departure end, which equates to a certain minimum climb gradient.  If I am lifting off with 1,500 feet remaining, I may need a steeper climb gradient to meet the same obstacle clearance expectations.  Similar to complying with an non-diverse runway ODP by making an earlier liftoff.

@Max Isn’t V2 typically lower than Vxse?  And thus you’d expect a higher V2 to clear a closer obstacle?

Could it have to do with the V1 policy your operator has adopted? Are you at Max FLEX in your scenario? EFB can unbalance for certain situations if your performance department has customised it accordingly, as far as I recall. In your case, it could be as simple as the EFB utilizing the whole runway to carry more energy even though minimum V2 is much lower, I guess.


text:
Low level orbit
What’s the lowest altitude for an orbit? Is there any limitation? I have looked in the internet for a reference but couldn’t find any.

This should be asked on Space.SE, rather than Aviation, although they might close it as a duplicate of one of these: https://space.stackexchange.com/q/66616/17995 https://space.stackexchange.com/q/41887/17995 https://space.stackexchange.com/q/50147/17995 https://space.stackexchange.com/q/48880/17995

This is more a question for space.se. However it has some aviation aspects too.
There is no limit per se for a satellite orbit. Air molecules are detectable very far from Earth surface, though their density continuously drops with altitude. The ISS orbits at about 400 km, but air is still present until the thermopause, that is an altitude of 500 to 1000 km.
Still, the practical low limit for orbiting seems to be 160 km.
Low Earth Orbit satellites
So most satellites have to oppose air drag of various intensity. Drag slows them down and decreases their orbital height. A satellite must be reboosted from time to time with an engine to re-accelerate the vehicle and rise it to a higher orbit. The required force is provided either by thrusters (jets) or ion engines. The reboost capacity is limited in time by the satellite design.
The lower the orbit, the more important the drag, the more frequent the reboosts. For a low orbit, either the quantity of fuel must be increased, which makes the launch more expensive, or the lifetime is reduced.
The Japanese satellite SLATS maintained a height as low as 167 km during 7 days using both thrusters and ion engines.
See also:

Very low Earth orbit (VLEO
How low can satellites go?

Note orbits are not necessarily circular, they can be eccentric, with a maximum and a minimum altitude. For an eccentric orbit, the satellite velocity varies along the path.
Kármán line
The practical limit for orbiting seems to be 160 km.
There is a reason, descending below requires a huge quantity of fuel for reboost, and the lifetime is very short. Still a low orbit can be useful for a short mission, e.g. with space-based radars, which power can be divided by 4 when the height is divided by 2.
In the other direction, the velocity of an aircraft must be increased as it flies higher, the reason is lift depends on the number of air molecules in contact with the wing per unit of time. Less air creates less lift unless the wing area is increased, but this has a limit, or velocity is increased.
Near altitudes of 90-115 km, an aircraft must fly at some velocity, and a satellite must orbit at about the same velocity to maintain a circular orbit and not fall. This is where flight in air and orbiting in space meet.
For this reason, the height of 100 km, named the Kármán line, is the "official" limit between atmosphere and space. More accurately, the height where flying with lift and orbiting require exactly the same velocity is 110 km, known as the maximum equilibrium velocity.
You can read more about this aspect in: Does Karman line depend on the wing area, and if not, why?
Is it easier to fly below 110 km than to orbit:

You can fly at velocities below the ones required for orbiting.

Materials used for the satellite must resist heating from drag, there is a practical limit.

Aerodynamics becomes a factor, the satellite design must be changed, this may not be compatible with the satellite functional requirements (e.g. antennas and solar arrays).

More fuel is required to reboost the satellite for a sustained orbit. There is also a practical limit.


text:
what's the difference between an ER and an LR variant of the 777-200
I was curious about the difference between the 777-200 LR and the 777-200 ER
ER stands for extended range
LR stands for long range
What's the between them difference? Why have two variants which do the same thing?
And if one has a longer range than the other, why didn't they just use that one instead of doing both the ER and the LR?
And any performance differences?(Takeoff Distance, Landing distance)

Somewhat related-- https://aviation.stackexchange.com/a/48466/34686 -- there are a lot of different ways to tweak a design and each choice is a trade-off of some kind--

The Boeing 777-200ER is an upgrade of the original -200 variant with higher fuel capacity and MTOW (maximum takeoff weight). It is otherwise quite similar to the original variant. The 777-200LR is however part of the second generation of the 777 (along with the 777-300ER and 777-200F), which received a new wing with ~4m larger wingspan and is exclusively powered by the second generation GE90 engine (the -200ER has the option to use the PW4000, RR Trent 800 or first generation GE90). This gives the -200LR a significantly higher range:

777-200ER
777-200LR

Introduced
1997
2006

Wingspan
60.93 m
64.8 m

MTOW
297 550 kg
347 815 kg

Max. Fuel
137 460 kg
145 538 kg

Range
7065 NM
8555 NM

(source: Wikipedia)
This makes the -200LR suitable for ultra-long haul flights, while the -200ER is more commonly used for shorter long haul flights:

The B-market 777-200ER ("ER" for Extended Range), originally known as the 777-200IGW (increased gross weight), has additional fuel capacity and an increased MTOW enabling transoceanic routes. [...] As of 2018, 338 examples of the -200ER are in airline service. It competed with the A340-300. Boeing proposes the 787-10 to replace it. [...]
The 777-200LR ("LR" for Long Range), the C-market model, entered service in 2006 as one of the longest-range commercial airliners. Boeing nicknamed it Worldliner as it can connect almost any two airports in the world, although it is still subject to ETOPS restrictions. It holds the world record for the longest nonstop flight by a commercial airliner. [...] The -200LR was intended for ultra long-haul routes such as Los Angeles to Singapore. [...] Airlines operated 50 of the -200LR variant as of 2018. [...] The closest competing aircraft from Airbus are the discontinued A340-500HGW and the current A350-900ULR.

(Wikipedia: Boeing 777)
The Boeing 777X will be the replacement for the 777-200LR and -300ER.

Had a pilot on the jumpseat years ago who had just finished the 777 type rating course at Boeing, who would soon be flying it for an airline. Asked him about it; he gushed about it being a wonderful airplane; it's only problem: "it holds waaaay too much gas!"


text:
How does a trailing edge flap move on an Airbus A320?

Airbus flaps move in flight direction [link]

I don't exactly understand how that works, do they move just 180 degrees (opposite direction) of the flight path or is it something else? And doesn't every airliner have flaps moving this way (opposite direction of flight path)

If you compare the picture of the B737 in the answer you linked with the one of an A320, you'll see that both set of flaps move in the same way while deploying (see arrow):

(B737 belly, source)

(A320 belly, source)
Most probably there's a mistake in that answer.


text:
Why do planes sometimes get a 'forward boost' when the reverse thrust is deactivated?
I have noticed that sometimes when you suddenly close reverse thrust you get an increase in airspeed or a 'forward push'. How does it happen? Is it more common on some aircraft types than the others? What type of reverse thrust (clamshell, target type, cold stream) are more prone to this problem? Is this necessarily dangerous?

How much thrust is the engine still producing when the reversers are suddenly closed?

Are you basing this on data, or the experience as a passenger?

I have played flight sims and noticed that, also this has happened in real life. So i want to know why

There could be 2 (or more) explanations for this... first is that the stowing of the thrust reverser is faster than the spool-down of the engine. The other that I can think of is that you aren't feeling an acceleration, just not as much deceleration, just like in a car if you brake hard then let off.

I've done that in CRJs when I was just freshly type rated until I learned to modulate the reversers during the landing (I don't remember the effect happening, or being noticeable at least, in the sim during training).
When you deploy the reversers you have a range where you can modulate the thrust between idle and max, just like in forward, but just using the reverser levers.  At the bottom of the range, there is a reverse idle setting just above fully stowed. So you can unlatch and pull the levers up a bit where you are at reverse idle (usually there is a positive detent to tell you you are in idle reverse), then you can move them farther up and thrust will increase until they are all the way and you are at max reverse thrust.  You can go back down and modulate the thrust the other way, down to idle reverse, with the reversers still deployed with the engines at idle, until you push them all the way back down to the stow position.
Technique wise, when you go to max reverse and you want to remove reverse thrust, you should ease the levers to idle reverse first, let the engines spool down, then move them to stow.
A lot of pilots will just go straight from max reverse directly to the stow position in one action, not pausing at idle reverse, because they may be a little busy with the landing, or just inexperienced.  The result is the engine's fan is still driving significant air, as it's spooling down, while the reverser doors or clamshells have stowed, resulting in a residual forward thrust kick from the fan's inertia that lasts a second or two.
It feels like you let off the brakes a bit then put them back on.


text:
Do symmetrical airfoils generate induced drag?
Cambered airfoils generate induced drag because they have a pressure differential between the upper and lower surface. However symmetrical airfoils don't. So does this mean that symmetrical airfoils have no induced drag?
Image source:https://en.m.wikipedia.org/wiki/File:Airfoil_camber.jpg

The airfoils alone don't - but the wings made with them do. Especially if they have finite span and produce lift.

It depends on the angle of attack. If it is non-zero there will still be a different airflow on both sides - leading to different pressures and thus induced drag.

If an airfoil is producing lift, then it will be producing induced drag.
Both cambered and symmetrical airfoils have an angle of attack at which they produce no lift, no induced drag, and no pressure difference between top and bottom. While this angle lines up nicely with the geometrical mid line of the symmetrical airfoil, it is offset from the apparent mid line of the cambered airfoil.
The cambered airfoil can produce more lift before the stall in the 'normal' lift direction, which is why it tends to be used for surfaces loaded in only one direction, like wings. A symmetrical airfoil would be used for control surfaces which might see equal loading in either direction.

Mmh, here there is a MISUNDERSTANDING OF TERMINOLOGY and even the wiki.en page about induced drag is a bit misleading.
Induced drag is a strictly 3D phenomenon, so speaking about induced drag for an airfoil is physically and mathematically not correct since airfoils are, by definition, 2D geometric shapes.
An airfoil generates lift, drag and moment, even if it is a simple 2D shape. But what is defined as "induced drag" is an additional contribution to the drag that appears only introducing the third dimension (it is a bit like when the flow becomes supersonic, then yet another source of drag - the wave drag - appears as well).
So, the exact geometry of the 2D airfoil is not important here, only the fact that the "wing ends" is involved in generating the induced drag.

Qualitatively the induced drag can be explained looking at what happen at the wing's tips i.e. when the wing ends. On the upper surface of the wing there is an underpressure while on the lower part of the wing there is an overpressure. At the tip, the underpressure over the wing sucks in the air which is pushed away from beneath the wing and "escapes" around the wingtips. This generates behind the wingtip a typical vortex as visible in the following picture:

B-727 in flight during vortex study with wingtip smoke generators. Source: https://www.dfrc.nasa.gov/Gallery/Photo/B-727/Large/ECN-3831.jpg
More correctly, a "sheet of vortex" is released from the trailing edge of the entire wing as visible in this standard representation:

Vortex sheet in the wake of a wing of finite span. Source: https://cdn-images-1.medium.com/max/1600/0*2gL-QOXqx51qNdAE.jpg
But normally only the vortex at the tip is taken into account for the sake of simplicity (even if this is not really correct). This vortex sheet locally decreases the angle of incidence therefore reducing the total lift generated by the wing. To compensate for this reduction, the angle of incidence has to be increased with a related increase of drag. This latter, and only this, is the induced drag. I perfectly agree that the naming is a bit misleading, but this, and only this, is the definition of induced drag.

I find https://en.wikipedia.org/wiki/Lift-induced_drag a way less misleading explanation than any talk about flow around the wing tips. While both are correct, just using different levels of description, talking about the flow around the wingtips leads people to think the induced drag can be significantly affected by changing the wingtips, while in fact changing the wing tip is at best equivalent to slightly increasing the wingspan. The angle of force and/or conservation of momentum and energy arguments don't have that problem, and immediately explain the inverse dependence on speed.

It is only a strictly 3D phenomenon according to the lifting line theory. Conservation of energy and momentum prove that any 2D section producing lift must also produce induced drag in an off itself without need to consider the wing tips.

@JanHudec: Using a 2D airfoil to describe a fundamental 3D phenomenon is physically and mathematically incorrect, even if Wikipedia says so
And yes, modifying the wingtip geometry is a useful (and used) way to control induced drag, together with other expedients like adjusting aspect ratio, sweep angle, taper ratio, t/c, twist angle and, obviously, airfoil geometry. In modern jetliner even the stiffness of the structure is "tailored" according to the different aerodynamic environments due to different loading conditions.

@JanHudec: Again, here there is a misunderstanding in terms. Navier–Stokes equations prove/model that an airfoil (or even a slice of a washing machine) generates lift, drag and moment, even in 2D. What is defined as "induced drag" is an additional contribution to the drag that appears only introducing the third dimension, so it is a strictly 3D phenomenon (that Prandtl smartly simplified via his lifting-line theory enough to make it solvable by hand)

Then there are two definitions that are clearly incompatible with each other. Because so far I always heard “induced drag” to mean the drag due to generation of lift (as opposed to friction and pressure drag that are there even if no lift is generated), but that is there even in 2D, even though yes, in 3D there is an additional contribution. Not a huge additional contribution though.

Then the other definition is wrong
Or actually only incomplete: the induced drag is the additional drag due to the generation of lift by a 3D body.
And the induced (aka 3D) drag is not small but actually the biggest source of drag for an airplane - at least until its minimum-power speed is reached, afterward the parasite drag takes over.

Would you believe I have a draft journal article about this. The use of the term "aerofoil" is a clear implication of 2-dimensional flow, or what Hurt [1] (and many others) call and infinite wing (AR = $\infty$). The tern wing is a clear implication for 3-dimensional flow. There is a clear logical distinction between how we teach these in aerospace (aeronautical) engineering and is best captured by Anderson [2]. That is, we teach 2D aerodynamics first, then we teach 3D aerodynamics.
Now, the aforementioned journal article I am working on is about confusions with this. In that, I have studied education literature on lift, and only one third of the time when an "aerofoil" is used and the implication is 2D flow, do they actually get this correct. For wings and 3D flow, this is much better, it is three quarters of the time when they are talking about wings and 3D flow that it is correct. So, as is reflected here in this discussion, there is a misunderstanding about the fundamentals of 2D flow.
As has been pointed out by others, there is no induced drag in 2D flow. Hence, the answer to the question is no, symmetric aerofoils, like all aerofoils, do not produce induce drag. Induce drag results from 3D effects, whereby energy is wasted by a wing imparting a downward velocity component to the air. This image from Hurt [1] is the best reference to this and is sadly never referred to once in 105 articles about lift education (or in books on the topic). As such, I recommend reading Hurt for the best description of induced drag. You will note in the image that the vertical velocity component for 2D flow returns to zero, while for 3D flow, it does not. Happy to share either some potential flow of CFD sims that support what was clearly well known in the 1960's.

1 Hurt, H. H. (1965). Aerodynamics for Naval Aviators. Aviation Supplies & Academics. PDF (25 MB)
2 Anderson, J. D. (2016). Fundamentals of Aerodynamics. McGraw-Hill Education. Publishers Link

Nice picture. This also clearly shows that a (2D) airfoil does not generate lift "pushing down" the air beyond it and receiving a kick back from Newton's third law

@sophit, just because the vertical velocity component for the flow does in fact become zero in the far field for a 2D flow around an aerofoil (again happy to share potential flow and CFD sims) does not mean there is no momentum transfer. This is why there is confusion, because people have become obsessed with statements of momentum transfer, even though every day in experiments aerodynamicists are busy measuring pressure to determine coefficients of lift.

Yes sure, lift, drag and moment are created but not because Newton's third law has pushed the airfoil upward

@sophit, it is an accepted fact that the streamlines in the far field return to horizontal, that is at an infinite distance away v_y = 0. Where lift can be expressed at the outlet boundary is in the divergence of the velocity vector. Note that Navier-Stokes requires conservation of a mass, and hence div(v) = 0. If the wake behind an aerofoil is slowed, that gives a -ve div, hence to balance this, the flow converges. The downwash is hidden in the asymmetry of the convergence, where flow above has a greater -v_y, while the flow below has a smaller +v_y.

Reading many answers here, it seems to be not so accepted

@sophit, well I am happy to explain to them why they are also mistaken. See: https://www.youtube.com/watch?v=QKCK4lJLQHU&t=1282s

Nice link, thanks for sharing

Doug's book is the best if you can get a copy.

Neither 2-D "flow", nor Infinite airfoils exist. They are mathematical abstractions, like straight lines, or points - simplifications designed, (in this case), to avoid answering the real-world aerodynamic behavior which is non-linear and hard. Golly, in "2-D" flow, What, exactly, is flowing!? ideal imaginary points?  Drawing conclusions from analysis of the mathematics for 2-D flows or "infinite" airfoils, and attempting to apply it to any real-world question is at best mis-leading, and at worst downright not even wrong.

@Aerospace Doctor, How can there be momentum transfer in a Hypothetical, imaginary, mathematical approximation like  2-D flow. Golly, in 2-D, nothing can be flowing, (it takes 3-D to hold any mass). Mass requires volume, no? How can there be momentum without mass? Your analogy exists only in your mathematics, and cannot exist in the real world.

@CharlesBretana you clearly have never done any measurement in a windtunnel. A wing that spans the entire test section is how we have experimentally measured the lift and drag for the countless NACA aerofoils we use. That is why we then correct for AR, given there is no span wise flow and no wingtip vortices.

Even better, here is the legend JD Anderson, Introduction to Flight:
5.5 INFINITE VERSUS FINITE WINGS
As stated in Sec. 5.4, the airfoil data in Appendix D were measured in low-speed
subsonic wind tunnels where the model wing spanned the test section from one
side wall to the other. In this fashion, the ft.ow sees essentially a wing with no
wingtips, i.e., the wing could in principle be stretching from plus infinity to minus
infinity in the spanwise direction... The flow about this wing varies only in the x
and y directions; for this reason the flow is called two-dimensional.

@AerospaceDoctor,  Please tell me you do not believe that infinite wings, or 2-D airfoils, actually exist! No matter what how you interpret wind tunnel results. The most egregious examples of foolishness come from so-called "experts" who believe that their ideas must be true because they produce consistent, reliable, or useful results.  Nothing in Science is demonstrably "True", no matter how successful it might be. Newton's theory of Gravity (the theory, not math), and Niels Bohr's Copenhagen Interpretation spring to mind. Incredibly successful, useful, and accurate, but demonstrably false.

One way that "Experts" come to grief is when they assume, because of positive, useful, consistent results from some interpretation they make, or from statistical analysis based on confidence levels based totally on correlation beyond some threshold, that a causation relationship must exist between independent factors, even when ordinary common sense should tell them otherwise... Like positing that an infinite wing exists, or that analysis of the math for a  2-D airfoil could produce anything more than narrow, tailored conclusions about real-world aerodynamics.

Sorry @CharlesBretana I will definitely note your comments about 2D flow, and ignore John D Anderson... You are right, all the NASA and other documents with airfoil data from wind tunnels are all wrong. Those fools believing their force physical balances in real wind tunnels showing an aerodynamic reaction with a lift component that was just coincidentally similar to that estimated by thin aerofoil theory. They clearly did not do a controlled experiment, where the only thing altered was the AoA resulting in an observable change in 2D lift. Just out of interest, what should their conclusion be?

@Aerospace,  Your comment only tells me that you didn't read what I said. Or, perhaps you are way too comfortable using the "straw-man" argument technique. I never said wind tunnel data was wrong. (except when interpreted or presented as being a 100% accurate representation or measurement of the real world). What I said is that many implications drawn from that data, and/or conclusions, especially when they involve causation, are wrong.

@Aerspace, and the comment about "Coincidentally similar" makes my point. I don't know for sure that these conclusions are wrong. But you can't know for sure that they are right. The FIRST lesson you should have learned before doing any science is that correlation does not imply causation. Two events could be caused by the same third unknown, invisible event. Did smoking cause cancer, or did cancer induce it's victims to smoke? well, we know the answer to a fairly high degree of confidence NOW, but any scientist who pretended to know the answer in 1950 was a fool.

... and, to answer your question. I don't know what all their conclusions were, so I can't answer the question with certitude. Certainly, ALL their conclusions should be subject to confirmation, and should still be treated with an open mind and some skepticism. But To what degree any conclusion is relied upon should be based on careful analysis of how well the experimental methodology mirrors the real world, and what (if any) assumptions and/or simplifications were made in the mathematical analysis.

My first post here was, in all honesty, just a quibble. I fully understand how the analysis of a hypothetical 2-D flow can be used to understand some aspects of real-world flow. One of the first things I had to in my course in Complex Analysis was to model 2-D  flow around a specified 2-D boundary in the complex plane.  I only posted because your answer seemed to imply that 2-D flow and or infinite wings are real, and that all analysis of their behavior is applicable to the real world.

@CharlesBretana luckily, I teach research methods, which includes true experimental designs and what is required to give them internal validity to claim causality. This is the problem in developmental designs like those used in smoking v cancer research. That lack of internal validity there is what means you can only CO relate the things. Having validity and controlling confounding effects, means you can claim causation. Wind tunnel experiments are reproducible, and there are established standards and practices. Therefore countless 1st years globally in aerospace can measure the same effect.

I am also not attacking the strawman, you are. I claimed in my post that induce drag does not apply to 2D flow. Then you attacked the strawman, you claimed 2D was a mathematical construct with no basis in reality. I then countered with an explanation that 2D in aerodynamics means in wind tunnel, with no spanwise or tip flow. You then attacked the strawman again my claiming the aforementioned body of aerodynamics knowledge is flawed because of a general statement that "experts" claim causation when they mean correlation. Do you see a trend here, I do, but I will hold off on proposing a theory.

Induced drag is just the component of any aerodynamic force which lies in the direction the aircraft or airfoil is traveling in (aligned with the relative wind). So of course it does. If it is producing any aerodynamic force, then there must be drag. Only if we arbitrarily separate the total drag into multiple pieces and call one of the pieces induced drag can we make this distinction.  How we use the words sometimes makes things unnecessarily confusing. Remember, we used to describe the forces generated by the airflow in the cavity between the F-14 engine nacelles as "Lift". Does that Lift generate Induced Drag?
How about the Supersonic Shock Wave Lift generated by the XB-70 when it lowered the wingtips? Much confusion is generated by using definitions designed to be useful in one context in areas where they are not applicable.
It could, (and doubtless will), be argued, (strictly based on arbitrary semantic definition), that unless there is Lift, there cannot be induced drag, but that argument is based on semantics, (that induced drag, by it's own name, is the drag induced by, or due to Lift). That definition however, although useful at times, is at best misleading.  It ignores the fact that Lift itself is a component of the same aerodynamic force - the component that lies normal to (perpendicular) to the aircraft/airfoil direction of motion or relative wind. Lift AND Induced drag are both just components of any aerodynamic force. One is not the cause of the other. Going strictly by this interpretation, If there is no "Lift", (because the total Aerodynamic force is itself aligned perfectly with the aircraft velocity vector), then it doesn't make sense to call it Induced Drag. It is, however, still there, and it is still the component of the force aligned with the velocity.


text:
Is a plane stalled during landing rollout?

IMAGE SOURCE: https://pilotinstitute.com/wp-content/uploads/2022/01/Parts-of-a-Wing.jpg
As the lift dumpers are extended on landing they cause the air to move away from the wing. Also as the flaps are extended fully the airflow further separates from the wing? Does this cause the wing to stall even if its at 0 AoA during landing rollout?

Stall for a wing is defined as:

a reduction in the lift coefficient generated by a foil as angle of
attack increases. This occurs when the critical angle of attack of
the foil is exceeded. (Wikipedia, citing from: Crane, Dale, Dictionary of Aeronautical Terms)

So to be sure we should do testing to find the wing’s critical AoA in this state. If it's greater than 0 then the answer is no. It even can be generating some lift at this moment and given the sufficient thrust and speed it can probably take off (not advised).
But if we think of stall as turbulent airflow above the wing, the area
behind the spoilers are definitely stalled but not the whole wing.

I don't think you can "think of stall as turbulent airflow above the wing".  For sure, that happens at the stall point, but stall is precisely defined as the AoA when lift starts decreasing.   The turbulence behind the spoilers indicates they might be "stalled", but I'd question whether there is even a concept of "stalled spoilers": do spoilers ever experience an increase in lift based on Aoa, such that it could decrease to even find the stall?

One way to look at it is, a stall requiring the exceedance of critical angle of attack, and the critical angle of attack not being exceeded in the landing, what has actually happened is the spoilers have transformed the wing from a wing into something else.
You could say that the lift dumpers transform the lift making airfoil into just a randomly shaped object, like a sofa or a row of deck chairs, that can't make lift, stalled or not..

One might even say that those “lift dumpers” deflect air upward adding downforce to the wing which increases the braking normal  force.

Technically, the fully deployed spoilers disrupt the airflow enough so that only minimal lift is generated by the wings. They actually create so much disruption, that regardless of AoA there will not be a coherent lift force generated at that speed. Effectively, that could indeed be described as an aerodynamic stall, at least by definition.
The difference is in common sense. In this case, it is not a negative effect but actually desired to do so. It negates the wing lift so that the full weight of the plane rests on the gear for maximum breaking effect.


text:
What is the green arrow on the 777's airspeed tape? Does it exist on other types as well?

Image source: fsnews.eu
I found this green arrow on the primary flight display (PFD) of a 777. What does it show? I'm assuming it's for showing acceleration and deceleration but I am not too sure. Is my assumption correct? Also, does it exist on other aircraft types?

@ymb1 According to the link you found, it's from a sim product, which is still in development. I guess they are just showing off what the PFD could display.

The green arrow is the speed trend vector showing predicted airspeed in 10 seconds:

2 Speed Trend Vector
Indicates predicted airspeed in ten seconds based on current acceleration or deceleration.

(Boeing 777 FCOMv2 10.10.4 - Flight Instruments, Displays - Controls and Indicators)
Other Boeing aircraft with a speed tape on the PFD (or ADI for early 757/767) have the same feature (I checked 737NG, 747-400, 747-8, 767 and 777 FCOMs). All have very similar descriptions, although the one for the 747-8 is more detailed:

Trend Indication
The tip of the arrow indicates predicted airspeed in ten seconds based on airspeed,
inertial data, and winds. Rapidly changing wind speed or direction can affect
predicted airspeed with no change to indicated airspeed.

Airbus aircraft also have a very similar feature, as already explained in this question: What scale is the speed trend line on a PFD's ASI?

The Garmin G3X has an airspeed trend line too.


text:
find the height of the aircraft
The elevation of Airport A is 150ft AMSL and Airport B is 500FT AMSL. The QNH at A is 1013hPa and at B is 998hPa. An aircraft is overhead A with 5000ft indicated on the altimeter with 1013hPa set on the subscale, and sets heading en-route directly to B. If the altimeter subscale and indicated altitude remain unchanged, what will the height of the aircraft be overhead B (Assume 1hPa = 30ft)

Welcome to the site. We don't mind helping with homework questions as long as someone shows that they have tried to solve it themselves and has run into a problem. Please edit and show your thinking and math.

Since this has been answered in the comments, here's a fuller explanation:
The altitude of the aircraft at A is 5000 ft. The air pressure falls as it travels to B, so to maintain the same pressure altitude it must descend. The QNH falls by 15Hpa, so working from the figures in the question the aircraft descends 450ft. So far, so good.
Note, however, that the questions asks for the height above Airfield B, not the altitude. Height is referenced from ground level, and B is at 500ft.
Thus, the height is 5000ft, minus 450ft due to pressure, minus 500ft due to airfield elevation.
The correct answer is 4050ft.


text:
How long does it take to complete a partial and full flap overspeed inspection?
I am trying to determine the cost of a flap overspeed inspection. Ignoring the cost of any parts. On average, how long does a flap overspeed inspection take? Or how long would a partial/full flap overspeed inspection take?

On an A380? Cub? Citation?

Small business jet lets say the size of a citation.

On a regional jet size airplane, about 1-2 manhours. On a large airliner, maybe one guy on each wing would take 1-2 hours, so say 4 manhours.
You're just inspecting for things that are bent or broken.
If the inspection procedure requires removal of fairings and panels, maybe add another hour.


text:
FAA FOQA Program
The FAA's Flight Operations Quality Assurance (FOQA) program is currently voluntary for some operators. The program allows flight data to be recorded, analyzed, and then submitted to the regulator. Additionally, operators with approved FOQA programs are protected from enforcement under 14 CFR 13.401. What are the pros & cons of joining the regulators FOQA program from the operator stand point?

The data collected allows your company to analyze how the flight are going.  It records numerous data points which allows the operator to analyze each phase of flight.

How much runway was required for takeoff?
Was there a crosswind on takeoff that wasn't corrected for by the pilots?
Did you drift off the centerline while on approach?
Were you 50FT AGL crossing the runway threshold?  Were you on the appropriate VREF speed?
Did you land in the touchdown zone?  When did you apply brakes?

The idea behind collecting and analyzing this data is to see how well operators are adhering to company SOPs, procedures and training.
To answer your question specifically.  Submitting this data to the regulator allows them to understand industry trends and develop training recommendations to reduce the risk in that phase of flight.  I don't know of any cons as the data cannot be used in any investigation.  I feel more data is great to allow better training standards and objective to be created to make pilots better.
I am not associated with C-FOQA or anything like this.
https://vimeo.com/146674763


text:
What ground speed reference is used for anti skid control?
I was reading about slip ratio controlled anti skid system and had this one particular question in my mind.
In addition to the wheel speed data, slip ratio controlled systems need a ground speed reference to estimate and control slip ratio.  Going through the relevant literature, I came across multiple methods (like unbraked nose wheel speed/ IRS etc.) for this. Which among these is used in modern aircraft?

The shortest answer is probably just the Inertial Reference System (IRS) sensors. But that made me think that, for braking, "ground speed" isn't as important as "acceleration/deceleration" forces.

@MarkHu, this is not true. To maximizing braking, you want to prevent skidding. This means that you want to have the wheel rotation equal to the ground speed. In the extreme case of skidding your wheel is not rotating, but you have a ground speed (and there is no effective braking). Comparing the two speeds gives much better control than responding to the measured deceleration.

On the Airbus A320, the ADIRU (Air Data Inertial Reference Unit) data is used for the reference speed of the anti-skid system:

The system compares the speed of each main gear wheel (given by a
tachometer) with the speed of the aircraft (reference speed). When the
speed of a wheel drops below 0.87 times the reference speed, the
system orders brake releasing in order to maintain the brake slip at
that value (best braking efficiency).
In normal operation, the BSCU determines the reference speed from the
horizontal acceleration furnished by ADIRU1, ADIRU2, or ADIRU3.
If all three ADIRUs fail, the reference speed equals the greater of
either main landing gear wheel speed. Deceleration is limited to 1.7
meters/second² (5.6 feet/second²).

(Airbus A320 FCOM - Landing Gear - Brakes and Anti-Skid - Principle)
The Boeing 737 FCOM does not contain all the details of the antiskid system. However, this presentation confirms that ADIRU data is also used on the 737:

Antiskid System
The antiskid system monitors wheel deceleration and controls the brake
metered pressure to prevent skid conditions. These are the antiskid
functions:

Skid control operates at more than eight knots to control each wheel deceleration during normal antiskid and both wheels on each
main landing gear during alternate antiskid.
Locked wheel protection compares wheel speeds more than 25 knots difference between the two inboard or the two outboard wheels and
releases brake pressure from the slower wheel. [...]
Hydroplane protection decreases wheel brake pressure to wheel 1 and 3 when ground speed is more than wheel speed uses the ADIRU
ground speed.

The answers would be incomplete without mentioning the 'original' ABS. It was used very widely on civil and military aircraft alike throughout the world until at least the 1970s.
The system employed a flywheel, connected to the wheel through a one-way clutch. When the wheel turns forward, it spins up the flywheel. If the wheel locks, the flywheel 'overtakes' it, spinning faster than the wheel. This releases the brakes. The system is completely self-contained; each braking wheel could have one.
In this system, the 'reference' is the recent speed of the same wheel. Of course, this assumes that the wheel must spin up before applying brakes. This is sometimes difficult to ensure, esp. on a wet runway.

Just for historical reference, the Space Shuttle antiskid system didn't use a ground speed reference, just the speed of the other main gear wheels.

Each main landing gear wheel has two speed
sensors that supply wheel rotational velocity
information to the skid control circuits in the
brake/skid control boxes. The velocity of each
wheel is continuously compared to the average
wheel velocity of all four wheels. Whenever the
wheel velocity of one wheel is 60 percent below
the average velocity of the four wheels, skid
control removes brake pressure from the slow
wheel until the velocity of that wheel increases
to an acceptable range.

Source: Shuttle Crew Operations Manual page 2-14.7


text:
Where is the ram air inlet duct on the A220 for the fuel inerting system? Is it the same duct as the air conditioing?
Title says it all... just trying to find out if there is a second NACA duct for the fuel inerting system on the A220, or if it's the same duct that is used for the air conditioning inlet.

The fuel inerting system on the A220 uses bleed air from the engines:

FUEL TANK INERTING SYSTEM (FTIS) – OVERVIEW
The Fuel Tank Inerting System (FTIS) generates nitrogen-enriched air and
distributes it into the air space in the fuel tanks to provide non-flammable air in the tanks.
The bleed air, cooled by a ram air heat exchanger, passes through an air
separation module that removes oxygen to create the nitrogen-enriched air.
The oxygen is expelled overboard and the nitrogen-enriched air is directed
into the fuel tanks.

(Bombardier CS300 FCOM 11-07 - Fuel Inerting System)
The air conditioning also uses that same bleed air. The NACA duct you mean is for the ram air used for cooling the bleed air, which is then dumped overboard again.
There are however NACA scoops for each surge tank on the A220:

In flight, NACA scoops located in each surge tank provide ram air
pressure to maintain a positive pressure on the fuel in the tanks. During
ground operations, NACA scoops provide static ventilation of the tanks
and relieve the buildup of air pressure during refueling or thermal
expansion of the fuel.

(Bombardier CS300 FCOM 11-02 - Fuel Storage System)

I find it amusing that you've illustrated an A220 answer with Bombardier CS300 images. I understand that the systems will, likely, be essentially identical, I just find it amusing. ;)

@FreeMan Well, the FCOM I have is an older one, which still says CS300, but it's the same aircraft.

Ah. I didn't realize that. It might be worth a note in the answer then.

Ah that makes sense. I had initially been confused by the "ram air" label in the first photo you posted, and was wondering where the outside air was coming from if there was no separate intake for it.
The same bleed air is cooled by ram air for both the air conditioning and the fuel inerting system.
Thanks!

Ref OP, there is a separate ram inlet for the Inerting system. The cooling ram inlet is positioned on the LHS near centerline of the belly fairing right next to the service access panel.


text:
What methods are used to calculate the geometry needed to obtain a bell-shaped lift distribution?
In aerodynamics, what methods do people use to calculate the geometry of a wing (and body) needed to obtain a bell shaped lift distribution? Please go into as much detail as possible for a stack exchange answer. Also, when feasible, please provide resources to back up claims.

@ymb1 I appreciate the references.

To whoever disliked this question: Could you please comment why you disliked it and give suggestions on improvements

For an unswept wing of an at least moderate aspect ratio in subsonic flow you can use the Multhopp panel tabulation. This divides the wing into an odd number of sections and is easily implemented on a computer.
Over the years, extensions have been published which allow to include wing sweep and the influence of fuselages. In general, any method based on potential flow theory will do.
The way to achieve a bell distribution is to start with an arbitrary wing design and then to iterate chord and twist in order to achieve the desired distribution for the desired angle of attack.

Thank you this is very helpful!

I think you are referring to a bell-shaped lift distribution for the entire aircraft.
If you take a look at the SU-27 fight jet, you can see that the wings aren't distinct from the center body of the jet. You see that wings merge together with the central body, creating one single lift body of the aircraft, as opposed to the traditional design where wings and the center of the aircraft.
This design has been proven to be extremely aerodynamic, and this aerodynamic configuration is still being used today on many fighter jets, examples include the Chinese J-11 and the Russian SU-35, which are all very good jets.

Moreover, if you are looking to reduce drag overall, you can also try to have winglets for your aircraft. They can reduce the amount of vortex drag created by your wings. But just be careful because winglets will reduce the maneuverability of your aircraft and will negatively impact supersonic flight performance.

Hopefully this is the answer you were looking for.
If not, have a nice day :)

Thank you for the answer. Although it is not exactly what I am looking for (due to ambiguity in the question which I will fix shortly), you did bring up a good point about including the body into my design. I will keep this in mind. Have a nice day as well! :)

No problem. And let me just add something to my answer. The reason why I mentioned the design of merging wings into the body is because this would allow the body to contribute to the lift as well. And thus, it would create a U-shaped lift distribution as the middle of the aircraft (body) would have the largest amount of lift, and the sides (wings) would have less.


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Why XFOIL predicts a lower drag coefficient at higher Re?
A well known idea is that for aerodynamic/streamlined bodies the friction drag is larger than pressure drag. Hence, a laminar boundary layer is preferable (only looking at minimal drag, no other design conditions).
Then, looking at the figure below it can be seen that the drag coefficient decreases with increasing Reynolds number, even at low angles of attack where one would not expect any significant flow separation. Also, up to Re 500000 one could expect laminar flow, but 'for sure' at Re 200000.
Is the XFOIL simulation realistic, or are there assumptions in the code which imply this behavior?

Why do you expect a drag coefficient increase? Lower $c_{\textrm{d}}$ at higher Re-Numbers is the expected behavior (e.g., see the figures for a sphere https://aviation.stackexchange.com/a/24755/8749  or a flat plate https://aviation.stackexchange.com/a/71276/8749)

Well, I am talking about the ratio of friction drag and pressure drag (assuming XFOIL does not use induced drag). Also, I am not talking about a strict increase over all values for the angle of attack.
But, given a low alpha where it is reasonable to assume that there is no flow separation (hence very little to no pressure drag) I expect the friction drag to be dominant. This is in turn lower for laminar flow i.e. Re < 5 * 10^5.  I expect a cross between Re 200000 and Re2000000 at an angle where flow separation occurs.

@Gypaets Also, case 2 in aviation.stackexchange.com/a/71276/8749 states that friction drag is dominant. Even more, at low Re i.e. laminar flow, the friction drag is lower. This is the reason for my question

Are you graphing Clift/Cdrag vs alpha, or just Cd vs alpha?  Assuming you are, as Velocity increases (more so than chord), your graphing would show these effects as Re increases.  My guess is we are looking at the same airfoil, sped up.  This would be a viscosity effect.

@RobertDiGiovanni You are correct. The y-axis label should be C_d only. I have plotted a NACA 63 421 airfoil.

@RobertDiGiovanni I still do not understand why the drag coefficient decreases? That is, assuming at low angle of attacks there is almost no pressure drag and friction drag is dominant. Therefore, low Re (laminar BL) should give lower Cd values

Did you carry out those calculations/simulations with XFOIL yourself? If you did, please check where the laminar-turbulent transition takes place. I am assuming that XFOIL determines the flow to be laminar for all Reynolds numbers for a pretty large part of the chord, so the high Re = low c_d relation will be dominant over higher Re => earlier transition => more drag.

Aerodynamic friction is caused by viscosity. The Reynolds number tells you how big viscosity is in relation to inertial forces. A bigger Reynolds number signifies lower viscosity. This means a higher Reynolds number almost always results in a lower friction coefficient. If you look at the plot below, the downward trend can be easily spotted.

Friction drag coefficient of a flat plate over Reynolds number (picture source). Note the double logarithmic axes.
If your airfoil is not designed to keep the boundary laminar, a higher Reynolds number always means less friction. For laminar airfoils at low angle of attack that special spot between laminar and turbulent flow can indeed mean a drag increase with higher Reynolds number, but that occurs normally only at the edges of the laminar drag bucket.
Within the laminar drag bucket and outside of it the laminar-turbulent transition is very gradual with a small forward shift of the transition point for an increase in the Reynolds number of several hundred thousand counts.

Excuse me for the (very) late reply; I disagree with your statement 'his means a higher Reynolds number almost always results in lower friction'. This is only true for the case where there is a turbulent boundary layer. My questions is aiming at why at low Re (assuming initial laminar BL) the Cd is higher using XFoil. See for example the difference in Cd at Re 5e5. Clearly, laminar is preferred over turbulent and is it in general not safe to assume that low Re have still laminar conditions?

@lWindy Guess why I used the "almost" weasel word: This is to account for the transition when laminar runs get shorter with Re and the drag coefficient within the laminar drag bucket increases with Re. But that happens only in a small AoA range and for Reynolds numbers between 100,000 and a few million. Your polar at Re = 0.55 million suffers from large separation bubbles at the transition point and will become better with turbulators, so the boundary layer is tripped a bit sooner.


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What are the limits for IFR legal heading indicator and compass?
I am a student pursuing CPL/MEIR at a flight school in Turkey that's equivalent of a part 141 in the US. Our instrument equipment check procedures for IFR flight contains comparing the compass heading with the heading indicator's heading. We have no six pack aircraft(G500 DV20'sand G1000 DA40NG's and DA42NG's) so we compare PFD heading with the compass heading and it is said that the difference must be +-6 degrees for IFR and +-10 degrees for VFR. I can't find the relevant regulation. Can somebody explain or at least point me in the right direction? Thank you so much. Fly safe.

The Earth constantly rotates at 15° per hour while the gyro is maintaining a position relative to space, thus causing an apparent drift in the displayed heading of 15° per hour. When using these instruments, it is standard practice to compare the heading indicated on the directional gyro with the magnetic compass at least every 15 minutes and to reset the heading as necessary to agree with the magnetic compass.
-- Instrument Flying Handbook FAA-H-8083-15B

The heading indicator should be checked at least every 15 minutes, and reset to the correct magnetic heading by reference to the magnetic compass (and compass correction card) during steady straight-and-level flight. For the heading indicator to be acceptable in normal operations, this correction should not exceed 3° in 15 minutes.
The Pilot’s Manual 3: Instrument Flying by Trevor Thom

The Pilot's Manual is not an official document, it is more along the lines of advice.
It is inevitable for the compass to differ from HI, the limit for 6 deg for IFR and 10 deg for VFR might be relative to a fix.(VOR, NDB)

It appeared to me that you were quoting the IFH and TPM, so I reformatted to make those more obvious quotes. If I've misinterpreted your intent, please [edit] to adjust.

Thank you for your time !


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Rating Title Differences with ICAO CPL
I graduated from an integrated ATPL programme in Turkey. Rating part of my CPL is comprised of two lines. Goes like:
IR/ME Valid until ...
MEP(land) Valid until..
However, my friend from my class(with the same exact training) got this CPL issued with only one line in the ratings part. Goes like:
MEP(land) IR Valid until ...
I think they will function the same but anybody have an idea on the style difference with these two CPL's and what could be the reason why?

Are the two dates the same in your case?

Nope they are different dates, but i also saw another friends licence with two lines with the same exact date.


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Hypothetical electrical failure in Aérospatiale SA 315B Lama in flight
What would happen to the functioning of the turboshaft engine of the Lama, if you have an electrical failure in flight?
A failure equivalent to turning the battery and generator off.

As stated "the functioning of the turboshaft engine"

Nothing will happen to the engine. You will probably lose engine indications and the ability to restart, but the engine will keep running.


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Can you choose to swap to a PPL when learning a (EASA) LAPL?
If I were to start flight training for a (EASA) LAPL and decide later (before completing the LAPL) that I wanted to do a PPL instead, is a switch possible?
Could the hours flown in pursuit of the LAPL count instead towards the PPL? Is there any penalty or things to consider when making this switch?
For context, I would be doing this training in France. I was meant to start doing a PPL with an instructor who is now having health issues, the only other teacher available can only teach LAPL.

There are organisations offering to upgrade LAPLs to PPLs, eg: https://highlandaviation.com/blogs/further-information/convert-lapl-to-ppl

A flight instructor in France let me know that the French Directorate General for Civil Aviation (DGAC) aims to not have student pilots who have committed to doing a LAPL swap to a PPL (but is okay with swapping from a PPL to a LAPL). How strictly the DGAC enforces this was not clear, but anecdotally, the flight instructor was aware of other flight clubs that had been contacted by the DGAC because of students swapping to PPLs.
In summary I believe the answer is that you may not swap to a PPL from a LAPL in France - however the certification you are doing does not need to be declared up front if the instructor teaches both PPL and LAPL.

Yes you can, and no , there is no penalty, at least not regarding the number of training hours. But be aware, that you need more training hours  for the PPL as for LAPL anyway: 45h vs. 30h. See https://leicesterairport.com/flight-training/ppl-lapl-comparison/ for a comparison.
However I would recommend complete the LAPL first. And only later, if you feel it's necessary, upgrading to PPL.


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Departing from closed traffic
I understand how to request closed traffic from the ground, and what it allows me to do (What does "closed traffic" mean?).
There's another closed traffic scenario that I imagine to be common. but I'm not quite sure how to string it all together.
Let's say I'm flying VFR to a Class D airport, where I'd like to do some touch-and-goes and then depart.
As discussed in How do I request a touch and go and then remain closed traffic?, I can request "closed traffic" on the way inbound, and let the tower know that I'll be departing (and in what direction).
On my last pass, I would announce midfield (e.g.) and expect "clear for the option". After acknowledging, should I also request/remind tower re:departure?
For example, should I acknowledge, "cleared for the option, NXXX123 ..." and then add, "... also, this will be my last pass, VFR departure to the north"? Should it be considered (and phrased) as a request?
Thanks!


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Are the airspace times of operation days of the week in local or Zulu time?
The margin of the Enroute High sectional H-4 claims that the times of operation of the airspace R-5107K near El Paso are "1400-0300Z, MON-FRI" (and other times by NOTAM).
If the days of the week are Zulu time then by converting this to local time would yield "0800-2100, MON-THU".
You could argue against this interpretation because usually, we use the days of the week to reference the starting time, which would mean that the airspace is available starting at "1400Z MON-FRI" which is still "0800 MON-FRI" when converted to local.
But that doesn't explain that on the Enroute Low sectional L-5 the OUTLAW MOA is available "0100-0500Z, MON-FRI by NOTAM".
If the days of the week are ZUlu time, converting that starting time to local time would yield "1800-2200, SUN-THU".
Assuming the NOTAM for this airspace is active, do they really mean that the airspace is active "1800-2200, SUN-THU"? or do they mean that it is active "1800-2200, MON-FRI"?

Related to: this and this

I don't have a reference to prove my point so I won't make this an answer, but I think that you are overthinking this... It should quite obvious to most that the intent is that the days of the week are the standard local workdays, but that the operating times are Z. Sure, plenty of businesses are open on weekends, and perhaps close on a Monday, but not so much the government.  To force people to interpret it in the way you are would be much too confusing.

On the VFR chart for OUTLAW MOA it lists the time of operation as "1800-2200, MON-FRI" which would imply that the days of the week being used for the IFR chart are the local days of the week. See this margin on the Phoenix VFR chart:

A different question answered that the FAA publishes these details of SUA in the JO 7400.10A "Special Use Airspace" and that all of the times in that document are local time.
If we look at OUTLAW MOA in this document, it also shows "1800-2200, MON-FRI".

I think that this is enough evidence for me to accept that the days of the week on the IFR charts are actually in local time even though the times of day are in Zulu time.

The VFR chart covering this area shows local times instead.
For this restricted area, it is 7 AM to 8 PM Mountain time MON-FRI.

That's different - those times are local, not Zulu.  The question asks about times that are explicitly marked Z, so this is irrelevant.

@TobySpeight The original poster was confused about the times. This is the same restricted area on a different chart. Restricted area R-5107 K is clearly open  7 AM to 8 PM Mountain time. Mountain time is -7 hours different from Zulu, so that is the offset that they need to use on the IFR chart.

Ah, I understand the relevance now.  I've proposed an edit which I think makes that clearer, and emphasises the days, since that's the focus of the question.  That said, guest was more interested in OUTLAW MOA, which changes day when converted between timezones.

@TobySpeight is correct, I'm much more interested in OUTLAW MOA specifically because of the day issue when changing time zones


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Can PC joystick simulate aerodynamic lock in spin?
I never explored joysticks for playing simulators on PC. Can some kind of joystick simulate aerodynamic lock during spin? You do hands off control stick and it stays there, won’t center as usual. I understand that force feedback is often included in such products, but can force feedback override auto centering?

I did play with force feedback joysticks but it was before I touch a real airplane's stick, so I cannot tell you how realistic it was but it definitely not autocentered when offline

Yes there are PC joysticks that can remain in an off center position: the active feedback joysticks, without the force feedback switched on. The stick feels pretty light in that situation and kind of remains where you put it, almost like a helicopter cyclic stick with the trim button released.
When switched on, the force feedback can act like a passive spring and accepts some input parameters for varying the feel. The neutral position can be varied, the return position of the spring. All pretty course, not bad for the money, although the stepper motor produces grainy feedback forces.

Part of the reason that there aren't many of these Force Feedback joysticks nowadays is that a company apparently had the patents locked up until just recently.  I don't know all the details, but it goes back to Sony v Immersion

I think it's more of a question about the simulator software than the joystick. For example Flight Simulator 2004 did have force feedback modeled into it, but I don't recall it being very sophisticated. I do not know the current status of "consumer grade" sim software, but I'm highly doubtful that they would be able to mimick actual feel of stick forces in a very natural way. Doing it would require quite a lot of computation.
FS2004 did emulate the affect of speed on control forces, but it was pretty crude.

Such effects can only be produced by so called 'force feedback' (FFB) devices. They exist, but few of them, if any (in the consumer market) produce satisfactory results.
Good force feedback simply cannot be done cheaply, and this restricts the market to niche enthusiasts and professionals. It is relatively easy to produce 'effects' such as vibration and impulses, and most consumer FFBs can do it. But to simulate the whole range of forces reqire a specialised drive and very solid construction.
In addition (and perhaps because of the small niche market of such devices), none of the common game simulators even attempt to simulate control forces reasonably. Tellingly, even the Windows interface (API) to drive FFB hasn't been updated for some 15-20 years (it even retarded a bit).
But all is not that bad. While I'm still not aware of any FFB stick/joystick controller that can do a satisfactory job, there are at least FFB yokes (such as this one) that can potentially deliver a relevant experience in GA aviation. They are not cheap by gaming standards ($1500+), but very cheap by aviation standards. To overcome software limitations, they typically extract flight data from the games and run their own force simulation, which requires some fiddling with set-up.
The car racing game industry is in a better position: the number of enthusiasts is larger, and an FFB steering wheel is a simpler device (being a single degree of freedom and lower force). There are quite a few racing FFB steering wheels on the market that can be considered decent, but even they are fairly expensive (~$1000).


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Is it easy for a Boeing 747 pilot to get a second typerating on the 777?
Is this easy and responsible? Since the 747 and 777 have a glass cockpit, you would think this would be easy.

"Easy" is a matter of opinion, and "responsible" is up to each individual. 747's are all cargo-ops now, so the only real reason a 747 pilot would want to add a 777 type rating is if they are moving to a new cargo assignment or they were an airline pilot that doesn't fly 747's anymore.

@ron beyer 747 is all cargo?  I am booked round trip on a BAW 747 round trip to London later this month.  -Skip Miller

@SkipMiller You're right... I was thinking that there were no more US operators and excluded my mind from the rest of the world...


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I have two VOR receivers, but if one VOR is out of tolerance, can I still fly IFR?
According to 91.171, I am not allowed to fly IFR unless:

"...unless the VOR equipment of that aircraft -

(1) Is maintained, checked, and inspected under an approved procedure; or
(2) Has been operationally checked within the preceding 30 days, and was found to be within the limits of the permissible indicated bearing error set forth in paragraph (b) or (c) of this section."

When it says "VOR equipment" am I to assume that requires ALL VOR receivers to be functional? Or is this assuming a single VOR receiver, therefore if one of mine is working, would I legally be allowed to fly?
Of course the 91.213 statement also says that "no person may take off an aircraft with inoperative instruments or equipment installed" and then goes on to list a variety of requirements to meet if it is to take off.
Thanks for the help!

There are a few things going on here:
First off you don't need any VOR receivers to fly IFR as per FAR 91.205 which lists out the minimum equipment for flying IFR.
Lets look a bit closer at 91.171

(a) No person may operate a civil aircraft under IFR using the VOR
system of radio navigation unless the VOR equipment of that aircraft -

So they only need to be in alignment if you are using the VOR system for your flight (which in many cases you will be). With that in mind, 91.171 does not specify how many receivers you need, simply that if you have one and want to use it, its aligned.
If you look closely at the clauses in 91.213 you will find that it deals  with minimum equipment lists (which your GA plane is unlikely to have). As such we should look at 91.213 (d)(2)(ii) and 91.213 (d)(2)(iii) which points us back to 91.205 and the "Kinds of Operations Equipment List for the kind of flight operation being conducted for which only a single VOR is required for IFR navigation if you intend to use the VOR system (as discussed above).
Keep in mind 91.213 (d)(3)(ii) indicates we must placard the inoperative or out of alignment unit

Deactivated and placarded “Inoperative.”

In some cases two VOR's may be required depending on what you want to do, and the approaches you intend on flying. There may be some things that are not available to you (or may not be practical) with only a single VOR receiver like the ILS/LOC into JFK's 13L which offers lower minimums if you can identify points of the LGA VOR (see UXHUB MINIMUMS).

Awesome, thanks Dave. Appreciate the thorough explanation. With 91.213 stating that "deactivation" is required, I'm curious how a VOR is deactivated without impacting your comms. (And yes, forgot to add that I am not req. to have a VOR for basic IFR day/night flight as per 91.205, was more focused on the VOR issue and its impact.)

@Tigre222 keep in mind you need the radio, antenna, and CDI for the whole shebang to work. Ive often seen CDI's throw flags or needles get stuck rendering it INOP even if the radio works. Depending on the radio the NAV section and COM sections may be independent internally and simply mounted in one housing.

Dave, good call. Definitely robust situations and lots to go offline. Appreciate it!


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At which stage of Aircraft Part 23 Type Certification needs flutter test / Vne 1.3x top speed?
Hello AviationExperts,
Recently Joby #eVTOL went down after trying to do extreme stress tests. Tracking websites have shown Joby Flying at 270 Miles per hour where as top speed of Joby S4 is 200 MPH
Can some expert decode why Joby is flying at 270 MPH ? At what stage of FAA certification warrants flutter test / Setting Vne ?
On a scale of 1 - 100 Part 23 certification where exactly they need to go for 1.3x top speed ?
Highly Appreciate Any Inputs
Thank you

What exactly applies depends on the date on which Joby applied for a type certificate. Before August 30, 2017, FAR23 consisted of a large body of specific regulations which had grown like weed over the preceding decades. Among those, FAR part 23.629 covered flutter, and paragraph (c) demanded that

Any rational analysis used to predict freedom from flutter, control reversal and divergence must cover all speeds up to 1.2 VD/1.2 MD, limited to Mach 1.0 for subsonic airplanes.

It seems Joby applied before that date so they need to prove compliance with 23.629(c). However, they should have stopped at 240 MPH, unless their dive speed is higher than 200 MPH. Top speed is not an exact wording - this could as well be the top cruise speed, and dive speed must be higher than that again. FAR part 23.335 requires dive speed to be 1.25 times cruise speed. This would bring you to the 270 MPH which caused their prototype to fail.


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Moment required for disk tilt Cierva Autogyro?
The first prototypes of the Cierva autogyro used a disk tilt mechanism rahter than cyclic pitch by means of a swashplate to achieve roll and pitch (DOI: 10.13140/RG.2.2.19168.94729).
Let's first assume the rotor blades are rigidly connected to the rotor hub and rotation axis. Moreover, let's assume we have two blades. Then, in order to tilt the rotor disk, one needs to apply a moment which scales with the angular momentum of the blades. Hence, the bigger the rotor and/or rotational velocity, the larger the moment required to tilt the disk.
Now, suppose a hinge is added whilst the blades remain rigidly connected to eachother like a seesaw. For the time being, assume the rotor spins in a room filled with stagnant air. What would the required moment be to tilt this rotor by means of disk tilt?
I am stuck on this conceptually: Suppose we move the rotor axis exactly and instantaneously when the rotor is positioned as in A. Since there is a hinge, nothing happens except the rotor will preces to such that the blades are perpendicular to the rotor axis.
Now suppose the instantaneous moment is applied when the rotor is at position B? Then, the blades act as a rigidly connected blade without hinge, right? Does the gyroscopic moment need to be overcome in this situation (omit the infinite moment for infinite acceleration)? Or are both situations the same and am I missing something?

The effect is identical to a situation where you hang the machine with its rotor head bolted rigidly to the ceiling of your garage, with the tiller of the rotor in your hand. You push the tiller to the side, and you find you are able to move the fuselage off the rotor's axis, say, 6 inches to a foot, before the effort gets too big.
So you have the ability, just by directly tilting the rotor off perpendicular to the fuselage, to move the C of G of the machine off axis some amount and impart a rolling or pitching moment as long as you maintain  the CofG offset from the rotor's axis some amount.
When you do it while flying, any displacement will get a roll or pitch response started, and just holding, say, 30 lbs of side force will keep the C of G offset from the rotor axis and maintain the roll at some value.  It's basically weight-shift control, like a hang glider.
Just about all small sporting gyros work this way, by tilting the rotor head directly, because the forces required a relatively low.
Once you get to a certain size, the forces just get too large and ponderous, and the only options are to hydraulically boost the tilting system (I don't know if that's ever been done) or articulate the rotor with a swash plate, by effectively making the blades themselves to the work.
Even with a articulating system, once you get to a certain size, you need to hydraulically boost that system as well to have manageable control forces, and most helicopters use boosted controls once the all up weight is in the 2000-3000 lbs range.

I think the example is a strange and counter-intuitive one. If I am not mistaken, what you're saying is that the rotor disk remains in the same position (as if it is hanging on a ceiling) and you push yourself and the aircraft using the stick connected to the rotor?
This implies that the force you need to apply is only related to the aircraft's weight. And what would happen when the rotor is spinning, but the fuselage is on the ground? Friction on the ground prevents the fuselage from moving, but I bet the rotor disk will. What moment is then required?

Go back to hang glider, or trike ultralight sitting on the ground on its gear. Next to it is a similar size autogyro, like the very early Benson with a tiller bar coming down from the rotor head. Tilt the wing on the trike by moving the control bar left, tilt the rotor on the gyro to the right by moving the control tiller to the left. Wing, spinning disc, same thing. In the air, same relationship but now you have the C of G suspended from the wing or rotor disc.Move the contro, tilt the wing/disc off the vertical axis, machine rolls.To maintain the roll, hold the input to keep the CG off axis.


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What is the direction of the centrifugal force of a hinged helicopter blade?
In helicopter dynamics the flapping angle and corresponding equation is a well known concept. Suppose now that an hinge is added to the rotor, as with gyrocopters and that there is an aerodynamic moment such that the flapping angle $\beta$ is formed.
In all literature, the centrifugal couple moment is countering the aerodynamic moment. However, why is the centrifugal force not directed radially outward from the blade sense (along $x_{img}$ axis). As this is the new rotary plane where the blade rotates over the $z_{img}$ axis.
My idea is that the centrifugal force is constructed from the rotational vector, and can be calculated in every coordinate system of choice. Therefore, the flap angle analysis can be made in the $x,y,z$ frame. Is this correct?
Moreover, does the use of a centrifugal force in the analysis require me to be in a rotary frame of reference? As my understanding is that this force appears when a transformation from a stationary to a rotary frame is made.
If that is the case, how then would the flapping equation look like when I stay in a stationary frame when there is no centrifugal force?

The centrifugal force is directed radially outward, but in the tip-path plane, itself defined by the combination of the rotating motion and the flapping oscillation.

...and the real axis of rotation is 'blown back' by the flapping oscillation, that real axis being perpendicular to the tip-path plane...

I don't get what's wrong with that picture? Centrifugal force is perpendicular with the axis of rotation, just like in the picture

Let me have a go at it. From https://en.wikipedia.org/wiki/Centrifugal_force, "the centrifugal force always points radially outward from the axis of rotation of the rotating frame". This also becomes apperent from the double cross product in $-m\vec{\omega} \times(\vec{\omega}\times \vec{r})$.
Therefore, one is allowed to choose any rotating coordinate system including a rotating coordinate system where the z axis corresponds with the physical rotor axis. The centrifugal force appears as a consequence of going from a stationary to rotating coordinate system. In this coordinate system, the resulting force (vector) is opposing the flap angle $\beta$.

You're right, I was assuming rotating already. I edited it

why is the centrifugal force not directed radially outward from the blade sense (along $x_{img}$ axis). As this is the new rotary plane where the blade rotates over the $z_{img}$ axis.

The centrifugal force is directed radially outwards from the axis of rotation, as also mentioned in @sophit's comment. But the centrifugal force is only one of the forces that will position the blade rotation plane and blade angle: the flight controls do a good job in tilting the rotation plane for flight path and airspeed control.
Some more explanations in this answer.

But is the axis of rotation defined as the physical axis i.e. coupled to the engine, or the imaginary axis which is $z_{img}$ in the figure?. This axis is the axis perpendicular to the tip path plane.

If it is the tip path plane, then I do not see why the centrifugal forces counters the blade flap. Namely, there is no arm and therefore no net moment around the $y, y_{img}$ axis.